For those who are interested, I've found the solution by making use of an alternative definition of a Runge-Kutta method.
We have two functions
$$
f_1(a,b) = -b \\
f_2(a,b) = \,\,\,\, a \\
$$
and the Runge-Kutta relation
$$
Y_i = y_n + h \sum_{i=1}^s a_{i,j} f(Y_i)
$$
By making use of this relation, we obtain
$$
\begin{cases}
A_1 = a_0 + h\left(-\frac{1}{4}B_1 -\left(\frac{1}{4}-\frac{\sqrt{3}}{6}\right)B_2\right) \\
B_1 = b_0 + h \left( \frac{1}{4}A_1 + \left( \frac{1}{4}-\frac{\sqrt{3}}{6}\right)A_2 \right) \\
A_2 = a_0 + h\left(-\left(\frac{1}{4}+\frac{\sqrt{3}}{6}\right)B_1 -\frac{1}{4}B_2\right) \\
B_2 = b_0 + h \left( \left(\frac{1}{4} + \frac{\sqrt{3}}{6}\right)A_1 + \frac{1}{4}A_2 \right)
\end{cases}
$$
This is a set of equations in function of $a_0$ and $b_0$, which we can solve by making use of computer software. Assume we have found $A_1$, $B_1$, $A_2$ and $B_2$ in terms of $a_0$ and $b_0$, then we can use the next equations
$$
a_{n+1} = a_n + h \left( b_1 \cdot f_1(A_1,B_1) + b_2 \cdot f_1(A_2, B_2) \right) \\
b_{n+1} = b_n + h \left( b_1 \cdot f_2(A_1,B_1) + b_2 \cdot f_2(A_2, B_2) \right) \\
$$
where the $b_1$'s on the right hand side are the Runge-Kutta coefficients ($b_1 = b_2 = \frac{1}{2}$)
Substitution the relations for $A_1$, $B_1$, $A_2$, $B_2$ and writing this out in matrix notation results in
$$
\begin{bmatrix}
a_{n+1} \\
b_{n+1}
\end{bmatrix}
=
\begin{bmatrix}
\frac{h^4-60h^2+144}{h^4+12h^2+144} & \frac{12h(h^2-12)}{h^4+12h^2+144} \\
-\frac{12h(h^2-12)}{h^4+12h^2+144} & \frac{h^4-60h^2+144}{h^4+12h^2+144}
\end{bmatrix}
\cdot
\begin{bmatrix}
a_n \\
b_n
\end{bmatrix}
$$
To use GF you need to multiply both sides by $z^k$ for some variable $|z|<1$ and sum over k. A generating function is a function of the type $G(z) = \sum_{k=0}^{\infty} a_k z^k$, so the first term on RHS will be $-4 G(z)$ and so on.
Algebraically you need to get $G(z)$ on LHS and some function $\Phi(z)$ on RHS and equate coefficients of $z^n$. This will be your 'closed-form expression' for $a_n$.
EDIT: here's the equation. You `massage' LHS a bit to get $\frac{G(z) -a_0 -a_1 z}{z^2}$ and set $2z=s$ and $ S=\sum_{k=0}^{\infty} ks^k $ to get
$$
G(z) -a_0 -a_1 z = -4 z^2 G(z) +8 z^2 S
$$
now you need to follow my suggestions above to get what you need
Best Answer
Let us start with a way to express the general solution:
$$\left(\begin{smallmatrix} a_1\\ a_{2}\\ a_{3}\\ .\\ .\\ .\\ a_{22}\\ a_{23}\\ a_{24} \end{smallmatrix}\right)=\underbrace{\left(\begin{smallmatrix}5\\ -1\\ a-5\\ -8.5\\ -2.5\\ 6\\ 7-a\\ 6\\ -2.5\\ -8.5\\ a-5\\ -1\\ 5\\ 6\\ -a\\ -1.5\\ -2.5\\ -1\\ a+2.5\\ -1\\ -2.5\\ -1.5\\ -a\\ 6\end{smallmatrix}\right)}_S+\alpha \underbrace{\left(\begin{smallmatrix} 1\\ 0\\ -1\\ 0\\ 1\\ 0\\ -1\\ 0\\ 1\\ 0\\ -1\\ 0\\ 1\\ 0\\ -1\\ 0\\ 1\\ 0\\ -1\\ 0\\ 1\\ 0\\ -1\\ 0\\ 1\end{smallmatrix}\right)}_{K_1}+\beta \underbrace{\left(\begin{smallmatrix} 0\\ 1\\ 0\\ -1\\ 0\\ 1\\ 0\\ -1\\ 0\\ 1\\ 0\\ -1\\ 0\\ 1\\ 0\\ -1\\ 0\\ 1\\ 0\\ -1\\ 0\\ 1\\ 0\\ -1\\ 0\end{smallmatrix}\right)}_{K_2} \tag{*}$$
with $a:=f(1)=\frac{5-7 \sqrt{2}}{2}\approx 2.4497$, for any $\alpha, \beta$.
Why considering 24 entries ? Because, for any $n$:
$$\sum_{k=n}^{k=n+23} f(k) = 0 \ \ \ \text{where} \ \ \ f(k):=5\cos \left(\frac{k\pi}{3}\right)-7\sin\left(\frac{k\pi}{4}\right)$$
(This could have been foreseen: $24$ is the shortest period for a cycle accomodating $\pi/3$ and $\pi/4$ like the Meton cycle for the Sun and the Moon...)
(See graphics below : the blue curve for function $f$ and the red curve for its cumulative values displaying the 24-periodicity),
As a consequence, we have the homogeneous recurrence relationship: $$a_{n+23}+a_{n+22}+2\left(\sum_{k=n+2}^{k=n+21} a_{k}\right)+a_{n+1}+a_n=0.$$
Let us stop here for a while. We know that the set of sequences $(b_n)$ (I voluntarily take a different notation) verifying relationship:
$$b_{n+23}+b_{n+22}+2\left(\sum_{k=n+2}^{k=n+21} b_{k}\right)+b_{n+1}+b_n=0.\tag{*}$$
is a vector space $V$ with dimension $24$. The set of sequences $a_n$ is therefore a subset (in fact an affine subspace) of $V$, with many constraints, all of them contributing to a final very low "dimensionality" of the issue.
Let us now switch to matrix algebra explanations.
Consider the $n= 24$ first linear relationships with a periodic indexing, meaning that $a_{25}=a_1, a_{26}=a_2$). This system of $n$ equations in $n$ unknowns
$$\underbrace{\left(\begin{smallmatrix} 1&0&1&0&0\cdots0&0&0\\ 0&1&0&1&0\cdots0&0&0\\ 0&0&1&0&1\cdots0&0&0\\ &.&.&.&.&.&\\ &.&.&.&.&.&\\ &.&.&.&.&.&\\ 0&0&0&0&0\cdots1&0&1\\ 1&0&0&0&0\cdots0&1&0\\ 0&1&0&0&0\cdots0&0&1\\ \end{smallmatrix}\right)}_A\left(\begin{smallmatrix} a_n\\ a_{n+1}\\ a_{n+2}\\ .\\ .\\ .\\ a_{n+21}\\ a_{n+22}\\ a_{n+23}\\ \end{smallmatrix}\right)=\left(\begin{smallmatrix} f(n)\\ f(n+1)\\ f(n+2)\\ .\\ .\\ .\\ f(n+21)\\ f(n+22)\\ f(n+23)\\ \end{smallmatrix}\right)\tag{2}$$
is not solvable because $A$ has rank $22$.
Indeed, the characteristic polynomial of $A$ is:
$$[x(x - 2)(x^2 - 2x + 2)(x^2 - 3x + 3)(x^2 - x + 1)(x^4 - 4x^3 + 5x^2 - 2x + 1)]^2$$
We can place the finger where the issue is: $A$ having $0$ as a double value, has a kernel with dimension $2$, meaning that there are in fact two arbitrary constants, which isn't surprizing... We must assign arbitrary values to $a_1$ and $a_2$ to get a solution $S$.
A basis of the kernel are the vectors $K_1, \ K_2$ with resp. entries $\cos k\pi/2$ and $\sin k\pi/2$, $\ k=1... 24$ (see formula (*).
Remark: I had investigated the 24 dimensional space of sequences. Here are some interesting facts, but in fact not at all useful for the solution of this problem.
The characteristic equation of (*) is
$$r^{22}(r+1)+2(\sum_{k=2}^{21} r^k)+(r+1)=0 \tag{1}$$
(which has the property to be reciprocal) has the following 23 distinct roots (obtained using a Computer Algebra System):
$$-1, \pm i, \pm w^k, k=1,2,...10 \ \ \text{where} \ w=e^{i \pi/11}$$
As a consequence, the general expression is:
$$a_n=K(-1)^n+L(i^n)+M(-i)^n+\sum_{k=1}^{10} C_k e^{ik \pi/11}+\sum_{k=1}^{10} D_k e^{-ik \pi/11}$$
for certain constants $K,L,M,C_k,D_k$ ($k=1:10$).
Due to the fact that the sequence $a_n$ is a real sequence i.e., such that $a_n=\overline{a_n}$, we can conclude that $L=M$ and for all $k$, $C_k=D_k$.
It remains to find unknowns $K,L,C_1,C_2,... C_{10}$ by setting the different constraints, which would lead to another much more complicated way to solve the issue.